The marketing division of a large firm has found that it can model the sales generated by an advertising campaign as \( S(u) = 0.75u + 1.7 \) million dollars when the firm invests \( u \) thousand dollars in advertising.

The firm plans to invest \( u(x) = -2.6x^2 + 52x + 200 \) thousand dollars each month, where \( x \) is the number of months after the beginning of the advertising campaign.

How quickly will the sales for this firm be changing when \( x = 17 \)?

To find the rate of change of sales when x = 17, substitute the given expression for u(x) into S(u) to get S(u(x)). Differentiate S(u(x)) using the chain rule and substitute x = 17 to find the rate of change.

Explanation:

To find how quickly the sales will be changing when x = 17, we need to find the derivative of the sales function S(u) with respect to x. First, we substitute the given expression for u(x) into S(u), which gives us S(u(x)). Then, we differentiate S(u(x)) with respect to x using the chain rule. Finally, we substitute x = 17 into the derived function to find the rate of change of sales at that point.

Using the chain rule, the derivative of S(u(x)) = 0.75[u(x)]' + 1.7, where [u(x)]' is the derivative of u(x). Differentiating u(x) = -2.6x² + 52x + 200, we get [u(x)]' = -5.2x + 52. Substituting these values into the derivative of S(u(x)), we get S'(u(x)) = 0.75(-5.2x + 52) + 1.7. Simplifying this expression, we get S'(u(x)) = -3.9x + 41.5. Finally, substituting x = 17 into S'(u(x)), we find that the sales will be changing at a rate of -3.9(17) + 41.5 = -7.3 million dollars per month when x = 17.

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